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HW 3 SOLUTIONS
The Way of Analysis
p. 54:
2.) Prove that every real number x has a unique cube root. This has two parts: show x
has at least one cube root and then show that y 3 = x and z 3 = x implies that y = z. The
first part just uses divide and conquer exactly as for square roots (pages 53–54).
Suppose now that y 3 = z 3 and y, z 6= 0 (it is easy to see that only 0 solves y 3 = 0). We
can factor this to get
(y − z)(y 2 + yz + z 2 ) = 0 .
Since
y 2 + yz + z 2 = (y + z)2 /2 + y 2 /2 + z 2 /2 ≥ y 2 /2 + z 2 /2 > 0 ,
we see that (y − z) = 0 as desired.
3.) Suppose that |xn | ≤ 2−n and yn =
N
X
i=j
|xn | ≤
N
X
Pn
i=1
xi . Prove that yn converges. Note that
2−j = 21−j − 2−N < 21−j .
i=j
It follows that for k > j > M we have (by the triangle inequality)
X
k
X
k
|yk − yj | = xn ≤
|xn | < 2−M .
n=j+1 n=j+1
Since this can be made arbitrarily small for M sufficiently large, we see that yn is Cauchy
and hence converges.
√
√
7.) Prove that a > b > 0 implies a > b:
Using the arithmetic rules for a field, we see that
√ √
√
√
a − b = ( a + b) ( a − b) .
√
√
√
√
Since √a and b are positive (by definition;
see
Theorem
2.3.3),
so
is
(
a
+
√
√ b). Therefore,
√
√
√
( a + b) has a positive inverse 1/( a + b). We can now write ( a − b) as the product
of two positive numbers:
√
√
√
√
( a − b) = (a − b) [1/( a + b)] .
10.) Prove that the irrationals Qc are dense in R; i.e., any x ∈ R can be represented
by a Cauchy sequence of√irrationals. Note that if√y ∈ Qc and z ∈ Q, then yz ∈ Qc (since
y = [z(yz)−1 ]−1 ). Since 2 ∈ Qc , it follows that 2/N ∈ Qc for every N ∈ N. Namely, we
get arbitrarily small irrational numbers.
Suppose now that xj is a Cauchy sequence of rationals for x. You should now try and
show that the sequence of irrationals
√
yj = xj + 2/j
1
2
HW 3 SOLUTIONS
then converges to x. (This is relatively straightforward, but should be included in your
solution.)
p. 84:
1.) Find the limit points:
• a) 1/n + (−1)n has limsup 1, liminf −1, and no other limit points (since 1/n goes to
zero).
• b) 1 + (−1)n /n has limit 1. (Hence this is the only limit point.)
• c) (−1)n + 1/n + 2 sin(nπ/2) has limit points 1 and −3. To see this, we can throw
away the 1/n since this goes to zero; then notice that 2 sin(nπ/2) is just
2, 0, −2, 0, 2, 0, −2, 0, . . .
2.) Suppose xn = yn + zn is bounded where yn increasing and zn decreasing. Does it
converge? Maybe not! For example, if yn = n and zn is the sequence
−1, −1, −3, −3, −5, −5, . . . ,
then yn + zn is the sequence
0, 1, 0, 1, 0, 1, . . . ,
However, if yn and zn are bounded, then they converge to real numbers (by our theorem
about monotone sequences) and hence so does xn (since the sum of convergent sequences
converges).
4.) Prove sup(A ∪ B) ≥ sup A and sup(A ∩ B) ≤ sup A. By definition, if y = sup(A ∪ B),
then y ≥ a for any a ∈ A ∪ B. In particular, y is an upper bound for A. Since sup A is the
least upper bound, we get
sup(A ∪ B) ≥ sup A .
The second claim follows from the first by writing A = (A ∩ B) ∪ A.
7.) Construct a sequence whose limit points are exactly the integers. To simplify notation,
we will do it for the natural numbers and leave the generalization to the integers for you to
work out. Here is the sequence:
1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5, . . .
First of all, this sequence contains every natural number infinitely often so automatically
has every natural number as a limit point. Second, since any two natural numbers differ by
distance at least one , any limit point must also be a natural number.
11.) Let aij be a rectangular array and consider the diagonal sequence
(D) a11 , a21 , a12 , a31 , a22 , a13 , . . .
Prove that any limit point of a row or column is a limit point of this sequence.
The cases of rows versus columns are symmetric, so we will just deal with one. Suppose
therefore that the i-th row has a convergent subsequence
(R) aij1 , aij2 , aij3 , . . .
with limit a. Note that the sequence jn is increasing and, therefore, so is the sum of the
indices i + jn . This means that aijn+1 occurs after aijn in (D) and therefore (R) is also a
subsequence of (D).
HW 3 SOLUTIONS
3
Do you necessarily get all limit points of (D) as limit points of a row or column? No!
Suppose that we had an array where aij = δij , i.e., where aij is one when i = j and zero
otherwise. In this case, the k-th row or column is zero after the k-th term, so 0 is the only
limit point of a row or column. However, infinitely many ones appear in (D), so one is a
limit point of (D).